Answer
$$\iint \limits_{R} x \sqrt{1-x^{2}} d A
=\frac {1}{3} $$
Work Step by Step
Given $$\iint \limits_{R} x \sqrt{1-x^{2}} d A, \quad R=[0,1] \times[2,3]$$
So, we have
\begin{aligned}
I&=\iint \limits_{R} x \sqrt{1-x^{2}} d A\\
&=\int_{2}^{3} \int_{0}^{1} x \sqrt{1-x^{2}} d x d y\\
&= \int_{0}^{1} x \sqrt{1-x^{2}} d x \int_{2}^{3} dy\\
&= -\frac {1}{2} \int_{0}^{1}(-2x) \sqrt{1-x^{2}} d x \ \ \ ( y|_{2}^{3}) \\
&= -\frac {1}{2} \int_{0}^{1}(- 2x) (1-x^{2})^\frac {1}{2} d x \ \ \ ( 3-2) \\
&= -\frac {1}{3} (1-x^{2})^\frac {3}{2}|_{0}^{1} \ \ \ \\
&= -\frac {1}{3} (1-1)^\frac {3}{2} +\frac {1}{3} (1-0)^\frac {3}{2} \\
&=\frac {1}{3}
\end{aligned}
